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# 5 8 m the object for the second lens is the image of

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Unformatted text preview: = 2 m − s1′ 1 11 =− s2′ f s2 1 inc re a s e s s1′ s 2 inc re a s e s 1 inc re a s e s s2′ P h y s ic s 2 1 2 Le c tu re 2 8 , S lid e 1 5 Multip le Le ns e s Exe rc is e s S up p o s e we no w d e c re a s e th e in itia l o b je c t d is ta nc e to 5 8 c m . Ap p lying th e le ns e q ua tio n, we fin d s 1’ = 2.48m Le ns s e p a ra tio n = 2 m What is the object distance s2 for lens 2? (A) s 1 = 5 8 c m f = 4 7 c m s 1 ’ = 2 .4 8 m s 2 = ­0 .4 8 m (B) s 2 = +0 .4 8 m (C ) s 2 = ­2 .4 8 m (D) s 2 = +2 .4 8 m (E) s 2 = +2 .5 8 m THE OBJECT FOR THE SECOND LENS IS THE IMAGE OF THE FIRST LENS s 2 = ­0 .4 8 OR s 2 = +0 .4 8 Im a g e o f firs t le ns is a VIR T UAL o b je c t fo r th e s e c o nd le ns P h y s ic s 2 1 2 Le c tu re 2 8 , S lid e 1 6 Multip le Le ns e s Exe rc is e s S up p o s e we no w d e c re a s e th e in itia l o b je c t d is ta nc e to 5 8 c m . Ap p lying th e le n s e q u a tio n , we fin d s 1’ = 2.48m Lens separation = 2 m s 1 =5 8 c m f=4 7 c m s 1 ’ = 2 .4 8 m s2 = ­0.48 m What is the nature of the FINAL image in terms of the ORIGINAL object? (A) R EAL UP R IG HT (B) R EAL INVER T ED fs2 s2 − f s 2 < 0 M2 = − s2′ s2 (D) VIR T UAL INVER T ED PICTURES EQUATIONS s2′ = (C ) VIR T UAL UP R IG HT Dra w R a y s a s a b o ve s 2 ’ > 0 re a l im a g e RESULTS s2’ = 0.24 m M2 > 0 M = M1 M2 < 0 M = ­2.1 inve rte d im a g e Physics 212 Lecture 28, Slide 17 Physics 212 Lecture 28, Slide Normal Eye Physics 212 Lecture 28, Slide 18 Physics 212 Lecture 28, Slide Checkpoint 2 A person with normal vision (near point 28 cm) is standing in front of a plane mirror. What is the closest distance to the mirror the person can stand and still see herself in focus? A. 14 cm B. 28 cm C. 56 cm Physics 212 Lecture 28, Slide 19 Physics 212 Lecture 28, Slide Checkpoint 2 A person with normal vision (near point 28 cm) is standing in front of a plane mirror. What is the closest distance to the mirror the person can stand and sti...
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